Let's walk through the proof based on the given image.

Given:

• $ABCD$ is a parallelogram with diagonal $BD$.
• $\triangle ABD$ and $\triangle CDB$ are the triangles to prove as congruent.

To Prove:

• $\triangle ABD \cong \triangle CDB$.

Hint:

• First prove the lines are parallel, then prove the triangles are congruent without using Theorem 32.

Proof (Two-Column Format)

Explanation:

• In this proof, we first identify that $ABCD$ is a parallelogram, which immediately gives us that opposite sides $AB \parallel CD$ and $AD \parallel BC$ are parallel.
• $BD$ is a common side (reflexive property), so both triangles share it.
• Since the opposite sides are parallel, the angles formed with diagonal $BD$ (angles $ABD$ and $CDB$, as well as $ADB$ and $CBD$) are congruent due to the alternate interior angles theorem.
• Using the ASA (Angle-Side-Angle) postulate, we conclude that $\triangle ABD \cong \triangle CDB$.

Do you want further clarification or more details on any of the steps? Here are 5 related questions to extend this topic:

1. How does the reflexive property help in proving congruence in triangles?
2. What other triangle congruence criteria could be used in different problems (e.g., SSS, SAS)?
3. Can you explain the significance of alternate interior angles in proving congruence?
4. What properties of parallelograms are useful in proving other geometric theorems?
5. How would the proof change if $ABCD$ were not a parallelogram?

Tip: Always look for shared sides or angles between triangles, as this often simplifies congruence proofs!